Question

For this problem, ignore air resistance.

You are to fire a rifle that shoots a particular bullet at 460 m/s at a target that is 45.7 m away and level with the rifle. Assuming that the acceleration due to gravity is 9.80 m/s², how high above the target must the barrel of the rifle be aimed such that the bullet hits the target?

* Physics - bobpursley, Wednesday, October 6, 2010 at 4:14pm

The real question is how far the bullet falls vertically during the trip.

timeinair=45.7m/460m/s

how far does an object fall in that time?

h=1/2 g t^2

* Physics - Steven, Wednesday, October 6, 2010 at 4:57pm

That's not entirely correct... That would only be true if the rifle was fired horizontally, giving the x component 460 m/s towards the target.

However, that 460 m/s has some UPWARD y component too, reducing the x component by some amount...

Thanks,
Steven

* Physics - Damon, Wednesday, October 6, 2010 at 5:06pm

Forget that upward component. Too fast to be significant. look at 460 cos (angle up)
the cosine of your angle up will be essentially 1 and the sine will be about 0

* Physics - Steven, Wednesday, October 6, 2010 at 8:37pm

I would like to forget the upward component, as the object IS in fact a bullet fired at a fast velocity from a gun towards a target not too far away.

However, we're graded on the correctness of our problem solving. And, additionally, this factor WOULD become an issue if we were firing a marshmallow gun or a pea shooter at this distance, or if you were firing a sniper rifle at a target over a mile away.

Thanks,
Steven

Answer 1

We know, R=(u²sin2∆)/g ,∆=angle of projection

So, 45.7=(460²sin2∆)/10
sin2∆≈1/460≈2∆ ,as ∆ is too small
∆≈1/920
So, tan∆≈∆ as ∆ is small
h/R≈∆
h≈45.7/920≈0.05m=5cm